Question: Let $f(x)=x^{-2}$. $f'(2)=$
Explanation: Let's first find the expression for $f'(x)$ and then evaluate it at $x=2$. The derivative of $f$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a negative number.) $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(x^{-2}\right) \\\\ &=-2x^{-2-1} \gray{\text{The power rule}} \\\\ &=-2x^{-3} \end{aligned}$ So we found that $f'(x)=-2x^{-3}$, which can also be written as $-\dfrac{2}{x^3}$. Now let's plug ${x=2}$ : $\begin{aligned} -\dfrac{2}{({2})^3}&=-\dfrac{2}{8} \\\\ &=-\dfrac14 \end{aligned}$ In conclusion, $f'(2)=-\dfrac14$.